我想在JavaScript中使用XMLHttpRequest发送一些数据。

假设我在HTML中有以下表单:

<form name="inputform" action="somewhere" method="post">
  <input type="hidden" value="person" name="user">
  <input type="hidden" value="password" name="pwd">
  <input type="hidden" value="place" name="organization">
  <input type="hidden" value="key" name="requiredkey">
</form>

如何在JavaScript中使用XMLHttpRequest编写等效内容?


当前回答

var util = {
    getAttribute: function (dom, attr) {
        if (dom.getAttribute !== undefined) {
            return dom.getAttribute(attr);
        } else if (dom[attr] !== undefined) {
            return dom[attr];
        } else {
            return null;
        }
    },
    addEvent: function (obj, evtName, func) {
        //Primero revisar attributos si existe o no.
        if (obj.addEventListener) {
            obj.addEventListener(evtName, func, false);

        } else if (obj.attachEvent) {
            obj.attachEvent(evtName, func);
        } else {
            if (this.getAttribute("on" + evtName) !== undefined) {
                obj["on" + evtName] = func;
            } else {
                obj[evtName] = func;
            }

        }

    },
    removeEvent: function (obj, evtName, func) {
        if (obj.removeEventListener) {
            obj.removeEventListener(evtName, func, false);
        } else if (obj.detachEvent) {
            obj.detachEvent(evtName, func);
        } else {
            if (this.getAttribute("on" + evtName) !== undefined) {
                obj["on" + evtName] = null;
            } else {
                obj[evtName] = null;
            }
        }

    },
    getAjaxObject: function () {
        var xhttp = null;
        //XDomainRequest
        if ("XMLHttpRequest" in window) {
            xhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        return xhttp;
    }

};

//START CODE HERE.

var xhr = util.getAjaxObject();

var isUpload = (xhr && ('upload' in xhr) && ('onprogress' in xhr.upload));

if (isUpload) {
    util.addEvent(xhr, "progress", xhrEvt.onProgress());
    util.addEvent(xhr, "loadstart", xhrEvt.onLoadStart);
    util.addEvent(xhr, "abort", xhrEvt.onAbort);
}

util.addEvent(xhr, "readystatechange", xhrEvt.ajaxOnReadyState);

var xhrEvt = {
    onProgress: function (e) {
        if (e.lengthComputable) {
            //Loaded bytes.
            var cLoaded = e.loaded;
        }
    },
    onLoadStart: function () {
    },
    onAbort: function () {
    },
    onReadyState: function () {
        var state = xhr.readyState;
        var httpStatus = xhr.status;

        if (state === 4 && httpStatus === 200) {
            //Completed success.
            var data = xhr.responseText;
        }

    }
};
//CONTINUE YOUR CODE HERE.
xhr.open('POST', 'mypage.php', true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');


if ('FormData' in window) {
    var formData = new FormData();
    formData.append("user", "aaaaa");
    formData.append("pass", "bbbbb");

    xhr.send(formData);

} else {

    xhr.send("?user=aaaaa&pass=bbbbb");
}

其他回答

下面的代码演示了如何做到这一点。

var http = new XMLHttpRequest();
var url = 'get_data.php';
var params = 'orem=ipsum&name=binny';
http.open('POST', url, true);

//Send the proper header information along with the request
http.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');

http.onreadystatechange = function() {//Call a function when the state changes.
    if(http.readyState == 4 && http.status == 200) {
        alert(http.responseText);
    }
}
http.send(params);

如果你有/创建一个对象,你可以用下面的代码把它转换成参数,即:

var params = new Object();
params.myparam1 = myval1;
params.myparam2 = myval2;

// Turn the data object into an array of URL-encoded key/value pairs.
let urlEncodedData = "", urlEncodedDataPairs = [], name;
for( name in params ) {
 urlEncodedDataPairs.push(encodeURIComponent(name)+'='+encodeURIComponent(params[name]));
}
var xhr = new XMLHttpRequest();
xhr.open('POST', 'somewhere', true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
xhr.onload = function () {
    // do something to response
    console.log(this.responseText);
};
xhr.send('user=person&pwd=password&organization=place&requiredkey=key');

或者如果你可以指望浏览器的支持,你可以使用FormData:

var data = new FormData();
data.append('user', 'person');
data.append('pwd', 'password');
data.append('organization', 'place');
data.append('requiredkey', 'key');

var xhr = new XMLHttpRequest();
xhr.open('POST', 'somewhere', true);
xhr.onload = function () {
    // do something to response
    console.log(this.responseText);
};
xhr.send(data);

这帮助我,因为我想只使用xmlHttpRequest和post对象作为表单数据:

function sendData(data) {
  var XHR = new XMLHttpRequest();
  var FD  = new FormData();

  // Push our data into our FormData object
  for(name in data) {
    FD.append(name, data[name]);
  }

  // Set up our request
  XHR.open('POST', 'https://example.com/cors.php');

  // Send our FormData object; HTTP headers are set automatically
  XHR.send(FD);
}

https://developer.mozilla.org/en-US/docs/Learn/HTML/Forms/Sending_forms_through_JavaScript

短而现代

您可以使用FormData捕获表单输入值并通过fetch发送它们

fetch(form.action, {method:'post', body: new FormData(form)});

function send() { let form = document.forms['inputform']; fetch(form.action, {method:'post', body: new FormData(form)}); } <form name="inputform" action="somewhere" method="post"> <input value="person" name="user"> <input type="hidden" value="password" name="pwd"> <input value="place" name="organization"> <input type="hidden" value="key" name="requiredkey"> </form> <!-- I remove type="hidden" for some inputs above only for show them --><br> Look: chrome console>network and click <button onclick="send()">send</button>

有一些重复的作品涉及到这一点,但没有人真正阐述它。我将借用公认答案的例子来说明

http.open('POST', url, true);
http.send('lorem=ipsum&name=binny');

为了说明,我过度简化了这一点(我使用http.onload(function(){})而不是那个答案的旧方法)。如果你按原样使用,你会发现你的服务器可能会将POST正文解释为字符串,而不是实际的key=value参数(即PHP不会显示任何$_POST变量)。你必须在http.send()之前传递表单头文件来获取它。

http.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');

如果您使用的是JSON而不是url编码的数据,则应该传递application/ JSON